Bar count is spaces plus one, and cover comes off twice
Two small mistakes account for almost every wrong rebar quantity, and both survive review because the answer they produce still looks reasonable.
The first is dividing the span by the spacing and calling that the number of bars. It is not — it is the number of spaces. A row of bars has one at each end of every space, the ends are shared, and one is left over at the finish. Ten feet at 12 inches on centre is ten spaces and eleven bars. It is exactly the fence post problem, and on a mat it costs you twice: one bar missing in each direction, and an unreinforced strip a full spacing wide sitting against two of the four edges — which is precisely where a slab cracks first.
The second is treating cover as a one-sided allowance. Cover is the clear distance from the face of the concrete to the surface of the bar, and there is a face on both sides. A 20 ft slab with 3 in of cover has bars 19.5 ft long, not 19.75 ft. And the loss happens twice more, in the other axis: the first and last bars also stand back from the edges, so the strip over which the bars are distributed shrinks by the same 6 inches. On a 4 ft square footing with 3 in of cover, that is 3.5 ft of usable width out of 4 — twelve percent of the footing gone before a single bar is counted.
Round the spaces up, never to nearest. Spacing on a drawing is a maximum, not a target. A 10.3 ft span at 12 in gives 10.3 spaces; taking 10 leaves the bars 12.4 in apart, which is outside the drawing. Taking 11 puts them at 11.2 in, which is inside it. The calculator above always rounds up and then reports the spacing you will actually get, so you can see how far it landed from the nominal figure.
Steel is bought by weight, and the bridge is 0.00617 d²
The yard sells rebar by the pound or the ton. Your drawing is in feet, inches and bar sizes. The number that crosses that gap is the mass per unit length, and it comes from one formula: kg per metre = 0.00617 × d², with d the bar diameter in millimetres. Steel has a density of 7,850 kg/m³ and the bar section is πd²/4; put those together with d in millimetres and the constant falls out at 0.0061654, which every standard rounds to 0.00617.
In the imperial system the same formula produces the familiar ASTM A615 table, because a US bar number is simply the diameter in eighths of an inch:
| Bar | Diameter | lb per ft | kg per m |
|---|---|---|---|
| #3 | 3/8 in · 9.525 mm | 0.376 | 0.560 |
| #4 | 1/2 in · 12.7 mm | 0.668 | 0.995 |
| #5 | 5/8 in · 15.875 mm | 1.043 | 1.555 |
| #6 | 3/4 in · 19.05 mm | 1.502 | 2.239 |
| #7 | 7/8 in · 22.225 mm | 2.044 | 3.047 |
| #8 | 1 in · 25.4 mm | 2.670 | 3.981 |
Notice how fast the weight climbs. Going from #4 to #5 is a 25 percent increase in diameter and a 56 percent increase in weight, because weight follows the square. That is why substituting a heavier bar "to be safe" is an expensive instinct, and why a quote that specifies only spacing and not bar size tells you almost nothing about what the steel will cost.
A useful sanity check once you have the total: divide the weight by the slab area. A residential slab-on-grade with a single mat usually lands somewhere between 1 and 3 lb per square foot. A suspended two-way slab with top and bottom mats runs considerably higher. If your number is an order of magnitude away from that band, the error is almost always a units slip or a missing mat, not the arithmetic.
Cover is a corrosion barrier, and it fails quietly
Cover is not a placement tolerance. It is the only thing standing between the steel and the outside world, and it does two jobs at once: it keeps the concrete's alkalinity around the bar so a passive oxide film stays intact, and it slows the diffusion of chlorides and carbon dioxide that would destroy that film.
When it fails, the failure is slow, invisible and then sudden. Carbonation or chloride ingress reaches the bar. The passive layer breaks down. The steel starts to rust, and rust occupies several times the volume of the metal it consumed. That expansion works outward against a thin skin of concrete which has no tensile capacity to resist it, so the cover cracks parallel to the bar, then spalls off in sheets. Now the bar is exposed directly and the process accelerates. By the time you can see a rust stain, section has already been lost.
ACI 318 sets minimum cover by exposure, not by preference:
| Condition | Minimum cover |
|---|---|
| Cast against and permanently exposed to earth | 3 in |
| Exposed to weather or earth, #6 and larger | 2 in |
| Exposed to weather or earth, #5 and smaller | 1 1/2 in |
| Slabs and walls not exposed, #11 and smaller | 3/4 in |
The most common way cover is lost has nothing to do with the drawing. The mat is laid on the ground, someone forgets the chairs or spaces them too far apart, the crew walks on it, and the steel ends up in the bottom half inch of the pour — or resting on the subgrade with no cover at all. Bar chairs at roughly 3 to 4 ft in both directions, and a foreman who checks the mat height immediately before concrete arrives, are worth more than any amount of extra steel. A bar sitting on the dirt is not reinforcement; it is a crack initiator with a rust problem.
Lap splices: the length is the easy part
Any run longer than the stock bar has to be spliced, and the usual splice is a lap: two bars laid side by side and tied, transferring force through the surrounding concrete rather than through the steel directly. For a Class B tension lap, 40 to 50 bar diameters is the practical range on ordinary work — 25 to 31 inches for a #5.
For quantities, the point is that a lap is duplicated steel. It is not a joint of zero length: it is an extra 30 inches of #5 bar for every splice in the job. A 50 ft run made from 20 ft stock needs three pieces and two laps, so it consumes not 50 ft but about 55 ft of bar. Multiply by 25 bar lines and you have quietly added 125 feet of steel that never appeared in an estimate built by dividing lengths. This calculator solves the splice count properly — pieces × stock ≥ run + (pieces − 1) × lap — because each lap eats into the reach of the bar that follows it.
The harder question is where. A lap splice works by developing bond either side of the joint, so it belongs where the bar is least stressed. In a simply supported slab, the bottom bars are working hardest at midspan, so splicing them there is the worst available choice; splices belong near the supports. Continuous top steel is the mirror image — it peaks over the supports, so its splices belong near midspan. Get this backwards and the quantity is right while the structure is not, which is the more expensive of the two mistakes. Stagger adjacent splices as well: putting every lap in a mat at the same station creates a plane of doubled bar and halved effective steel.
Opening up the spacing is a structural decision, not a saving
Spacing is the input people adjust when a quantity comes back higher than expected, and it is the one input on the drawing that is least available for adjustment.
A mat resists cracking by being close enough to the crack that it can hold it shut. Widen the spacing and each bar carries more force, so it stretches more before it takes load, and cracks open wider before the steel does anything about them. Wider cracks let in water and chloride, which brings you back to the corrosion problem from the other direction. Codes cap slab bar spacing for exactly this reason, typically at the lesser of about three times the slab thickness or 18 inches, with tighter limits for crack control in exposed work.
Going from 12 in to 16 in centres removes about a quarter of the steel and roughly a quarter of the steel cost. On a residential slab that might be a few hundred dollars, against a slab that now cracks in a pattern you will be looking at for thirty years and cannot repair without demolition. If the quantity is genuinely too high, the productive conversation is with whoever produced the drawing about bar size and mat layout, not with the tape measure on site.
Tie wire, chairs, and the mat that ends up on the ground
Ties do not carry structural load. They hold the mat in the position the drawing specifies until the concrete sets, which turns out to be the whole job.
The count is a product, not a sum, and this is where estimates go wrong by an order of magnitude. Eleven bars one way crossing twenty-one the other produce 231 intersections — not 32. At about a foot of wire per tie, that is 231 ft of wire for one small slab. Sixteen gauge annealed wire runs roughly 250 ft to the pound, so budget by the roll and not by the handful.
Tying every intersection is standard for mats that will be walked on, for anything vertical, and for edges and corners. On large flat areas a staggered pattern tying every other crossing is common practice and roughly halves both the wire and the labour without letting the mat wander. What is not optional is support: chairs, bolsters or dobies at 3 to 4 ft in each direction, sized for the cover you specified. Skimp there and the entire cover calculation becomes fiction the moment somebody steps on the mat.
Checking a rebar quote before you sign it
Three checks catch most of what goes wrong, and each takes under a minute.
- Recount one direction by hand. Take the slab dimension, subtract twice the cover, divide by the spacing, round up, add one. If the quote is one bar short in each direction, the estimator divided and stopped.
- Check the weight against the length. Total feet × the lb-per-foot figure for the bar size should reproduce the quoted weight within a few percent. A gap of ten percent or more usually means laps were omitted — or double-counted.
- Check the ratio. Weight ÷ area gives pounds per square foot. Compare it against similar work. A number far outside the usual band points at a missing mat, a wrong bar size, or a unit mix-up between metric and imperial.
And keep the two numbers separate in your head: the steel that ends up in the slab, and the steel you have to buy. They are not the same, because bars come in fixed lengths and cutting leaves remnants. Ten 13 ft pieces cut from 20 ft stock take ten bars, not six and a half — a 7 ft remnant is not half of another 13 ft piece. The calculator above keeps those two columns apart on purpose.
Everything stays in your browser
The whole calculation is arithmetic running on your device. Nothing is uploaded, nothing is stored and there is no account. Your dimensions, your supplier's price per pound and your margins are not our business.
Frequently asked questions
How many rebar do I need for a 20 by 10 ft slab at 12 in on centre?
With 3 in of cover, the mat is 19.5 ft by 9.5 ft. Across the 9.5 ft direction, 9.5 ft ÷ 1 ft gives 9.5, which rounds up to 10 spaces, so you need 11 bars running lengthwise, each 19.5 ft long. Along the 19.5 ft direction, 19.5 ÷ 1 = 19.5, rounds up to 20 spaces, so 21 bars running widthwise, each 9.5 ft long. That is 11 × 19.5 + 21 × 9.5 = 414 ft of #4 bar, which at 0.668 lb per foot is about 277 lb, or 22 stock bars of 20 ft. Divide by spacing and you get 9 and 19 — two bars short, with an unreinforced strip along two edges.
How do you convert rebar length into weight?
Multiply the length by the mass per unit length, which is 0.00617 × d² kilograms per metre with the diameter d in millimetres. A 12.5 mm bar is 0.00617 × 156.25 = 0.964 kg/m; a 10 mm bar is 0.617 kg/m. In imperial terms the same formula gives 0.668 lb/ft for a #4 bar, 1.043 lb/ft for #5 and 2.670 lb/ft for #8 — the ASTM A615 table values. This matters because steel is sold, quoted and invoiced by weight. A calculator that stops at total length has stopped one step before the number you actually need.
How long does a rebar lap splice have to be?
For a Class B tension splice, 40 to 50 bar diameters is the working range used on most residential and light commercial drawings, so a #5 bar laps roughly 25 to 31 inches. ACI 318 computes the real figure from concrete strength, bar size, cover, spacing, epoxy coating and whether the bar is a top bar, and it can land well outside that range. What matters for quantities is that the lap is duplicated steel: two bars running side by side for the lap length. Ignoring it always under-orders, and the shortfall grows with every splice in the job.
How much tie wire does a rebar mat use?
One tie per intersection, and the number of intersections is the product of the two bar counts, not their sum. A mat with 11 bars one way and 21 the other has 231 intersections, not 32. At about 12 inches of wire per tie that is 231 ft of wire — around 2.5 lb of 16 gauge. Tying every other intersection in a staggered pattern is common on flat slab work and roughly halves it; edges, corners and anywhere the mat is walked on still get tied every crossing.
Do I need to add waste to a rebar order?
Yes, 5 to 10 percent. Bars come in fixed stock lengths, so every cut leaves a remnant that may be too short to use anywhere else on the job. Add miscuts, bars bent during handling, and the ones that get damaged on site. This calculator separates the two: it reports the exact steel in the slab, the waste allowance on top of it, and how many whole stock bars that becomes once the cutting yield is taken into account — because ten 13 ft pieces do not come out of a 20 ft bar at 6.5 bars, they come out at 10.